The point (2t² + 2t + 4, t² + t + 1) lies on the line x + 2y = 1 for

The point (2t² + 2t + 4, t² + t + 1) lies on the line x + 2y = 1 

if (2t² + 2t + 4) + 2(t² + t + 1) = 1
i.e., 4t² + 4t + 5 = 0
Here, discriminent = 16 – 4 × 4 × 5 = –64 < 0.
$\therefore$No real value of t is possible.
Hence, the given point cannot lie on the line.